Finding the sum of n terms $S_n$ starting from sigma $k=0$

Question

$$\sum_{k=0}^{n} ((4k-3)\cdot 2^k)+4=(2^{n+3}+4)n-7\cdot2^{n+1}+15$$

How? I've tried everything but i don't see it. Any equivalent solutions are also welcome, thanks.

Answer 1

Use this important geometric sum :

Lemma 1 $$A(n)=\sum_{k=0}^{n} 2^k=2^{n+1}-1$$

We'll now find another sum :

Lemma 2

$$B(n)=\sum_{k=0}^{n} k 2^k$$

To find this sum simply break it in sums you already know from Lemma 1 like this :

$$B(n)=0+1 \cdot 2^1+2 \cdot 2^2+\ldots n \cdot 2^n=(2^1+2^2+\ldots +2^n)+(2^2+2^3+\ldots +2^n)+\ldots+(2^n)=2A(n-1)+2^2A(n-2)+\ldots +2^nA(0)=2(2^n-1)+2^2(2^{n-1}-1)+\ldots +(2^{n+1}-2^n)=n2^{n+1}-(2+2^2+\ldots+2^n)=n2^{n+1}-(2^{n+1}-2)=(n-1)2^{n+1}+2$$

Now your sum is just $$4B(n)-3A(n)+4(n+1)$$ which you can find by plugging in from Lemma 1 and 2.

Answer 2

$$\sum_{k=0}^{n} \{((4k-3)\cdot 2^k)+4\}$$ $$=-3(2^{n+1}-1)+4(2^{n+1}-2)+...+4(2^{n+1}-2^n)$$ $$=-3(2^{n+1}-1)+n \cdot 4 \cdot 2^{n+1}-4(2^{n+1}-2)+4(n+1)$$ $$=(2^{n+3}+4)n-7\cdot2^{n+1}+15$$

Answer 3

You can exploit a telescoping sum: $$ \sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)=(n+1)2^{n+1}, $$ because you also have: $$ \sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)= \sum_{k=0}^n\bigl(k2^k-2^{k+1}\bigr) $$ so that $$ \sum_{k=0}^n k2^k= \sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)+\sum_{k=0}^n2^{k+1} =(n+1)2^{n+1}+2(2^{n+1}-1) $$